# Variance in central limit theorem

Why is it that $\sqrt{n}(X_{n}-\mu)$ converges in distribution to $N(0,\sigma^{2})$ but $\sqrt{n}(X_{n}-\mu)/\sigma$ converges in distribution to $N(0,1)$?

• Shouldn't it be $\sqrt{n}(X_{n}-\mu)/\sigma^2$? – gung - Reinstate Monica Oct 2 '14 at 17:06
• Because $\text{var}(kX) = k^2 \text{var}(X)$ ? – Glen_b -Reinstate Monica Oct 2 '14 at 17:12
• I think you mean $\bar X_n$ (i.e., the sample mean), not $X_n$ (the $n$th observation – Russ Lenth Oct 2 '14 at 17:17
• @gung, no, I believe it the denominator is correct as stated – Russ Lenth Oct 2 '14 at 17:19
• True. My comment was based on the most conventional notations, and imputing the independence assumption since that is an instance of when the given statements are true. – Russ Lenth Oct 2 '14 at 18:33

For instance if $X_n$ is measured in meters, use a rod $\sigma$ meters long to measure $X_n$. Let's call the new unit an "sm." That changes $\sqrt{n}(X_n-\mu)$ meters into $\sqrt{n}(X_n-\mu)/\sigma$ sms. The former converges to $N(0\text{ meter},\sigma^2\text{ meter}^2)$ which, in the new measurement system, is $N(0\text{ sm},1\text{ sm}^2)$. Therefore $\sqrt{n}(X_n-\mu)/\sigma$ must converge to $N(0,1)$.